Are $a=1$ and $b=2$ the only positive coprime integers with $a\neq b$, such that
$$a^{p} + b^{p} \mid (a+b)^{p}$$
for some odd prime $p$ ?
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4$\begingroup$ The formulation of this question makes one wonder whether this is some competition problem. If you want an answer, you should add some background to the question! $\endgroup$– Jan-Christoph Schlage-PuchtaFeb 10, 2017 at 21:26
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$\begingroup$ Yes, please explain from the beginning ... and well-form your question ! $\endgroup$– Duchamp Gérard H. E.Feb 11, 2017 at 3:12
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1 Answer
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Zsigmondy theorem states that for any $p>1$, $a^p+b^p$ has a primitive prime factor, except when $\{a,b\}=\{1,2\}$. Such prime does not divide $a+b$. Hence, $a^p+b^p$ cannot divide $(a+b)^p$ unless it's the exceptional case.
answered Feb 11, 2017 at 2:46